Q. 185.0( 5 Votes )

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Answer :

Let p(x) = x3 – 3√5x2 – 5x + 15√5


As p(x) is cubic polynomial it will have 3 zeroes


As (x – √5) is a factor of p(x), p(√5) will be 0 by remainder theorem hence one zero of p(x) is √5


Divide p(x) by (x – √5)



Using dividend = divisor × quotient + remainder


p(x) = (x – √5)(x2 – 2√5x – 15) + 0


Remaining two zeroes of p(x) we can get by equating x2 – 2√5x – 15 = 0


x2 – 2√5x – 15 = 0


x2 – 3√5x + √5x – 15 = 0


x(x – 3√5) + √5(x – 3√5) = 0


(x + √5)(x – 3√5) = 0


x + √5 = 0 and x – 3√5 = 0


x = –√5 and x = 3√5


Hence all zeroes of x3 – 3√5x2 – 5x + 15√5 are x = √5, x = –√5 and x = 3√5


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